copyright seventh edition and expanded seventh …€¦ · · 2009-01-05copyright © 2005 pearson...

Copyright © 2005 Pearson Education, Inc.

Chapter 7

Systems of Linear Equations and Inequalities


7.1

Systems of Linear Equations

4Copyright © 2005 Pearson Education, Inc.

System of Linear Equations

A solution to a system of equations is the ordered pair or pairs that satisfy all equations in the system. A system of linear equations may have exactly one solution, no solution, or infinitely many solutions.


Example: Is the Ordered Pair a Solution?

Determine which of the ordered pairs is a solution to the following system of linear equations.

3x – y = 24x + y = 5(2, 4)(1, 1)


Example: Is the Ordered Pair a Solution? continuedSolution:

3x – y = 2 4x + y = 53(2) – 4 = 2 4(2) + 1 = 5

2 = 2 True 9 = 5 False

Since (2, 4) does not satisfy both equations, it is not a solution to the system.


Example: Is the Ordered Pair a Solution? continued

3x – y = 2 4x + y = 53(1) – 1 = 2 4(1) + 1 = 5

2 = 2 True 5 = 5 False

Since (1,1) satisfies both equations, it is a solution to the system.


Procedure for Solving a System of Equations by Graphing

Determine three ordered pairs that satisfy each equation.Plot the ordered pairs and sketch the graphs of both equations on the same axes.The coordinates of the point or points of intersection of the graphs are the solution or solutions to the system of equations.


Graphing Linear Equations

When graphing linear equations, three solutions are possible:

The two lines may intersect at one point, producing a system with one solution.

A system that has one solution is called a consistentsystem of equations.

The two lines may be parallel, producing a system with no solutions.

A system with no solutions is called an inconsistentsystem.


Graphing Linear Equations continued

The two equations may represent the same line, producing a system with infinite number of solutions.

A system with an infinite number of solutions is called a dependent system.


Example

Find the solution to the following system of equations graphically.

x – 3y = 92x – 6y = 3

Solution: Since the two lines are parallel, they do not intersect, therefore, the system has no solution.


7.2

Solving Systems of Equations by the Substitution and

Addition Methods


Procedure for Solving a System of Equations Using the Substitution Method

Solve one of the equations for one of the variables. If possible, solve for a variable with a coefficient of 1. Substitute the expression found in step 1 into another equation. Solve the equation found in step 2 for the variable.Substitute the value found in step 3 into the equation, rewritten in step 1, and solve for the remaining variable.


Example: Substitution Method

Solve the following system of equations by substitution.

5x − y = 54x − y = 3

Solve the first equation for y.5x − y = 5

y = 5x − 5

Then we substitute 5x − 5 for y in the other equation to give an equation in one variable.

4x − y = 34x − (5x − 5) = 3


Example: Substitution Method continued

Now solve the equation for x.

4x − 5x + 5 = 3-x + 5 = 3

-x = -2 x = 2

Substitute x = 2 in the equation solved for y and determine the y value.

y = 5x − 5y = 5(2) – 5y = 10 – 5 y = 5

Thus, the solution is the ordered pair (2, 5).


Addition Method

If neither of the equations in a system of linear equations has a variable with the coefficient of 1, it is generally easier to solve the system by using the addition (or elimination ) method.To use this method, it is necessary to obtain two equations whose sum will be a single equation containing only one variable.


Procedure for Solving a System of Equations by the Addition Method

If necessary, rewrite the equations so that the variables appear on one side of the equal sign and the constant appears on the other side of the equal sign. If necessary, multiply one or both equations by a constant(s) so that when you add the equations, the result will be an equation containing only one variable.Add the equations to obtain a single equation in one variable.


Procedure for Solving a System of Equations by the Addition Method continued

Solve the equation in step 3 for the variable.Substitute the value found in step 4 into either of the original equations and solve for the other variable.


Example: Multiplying Both Equations

Solve the system using the elimination method.6x + 2y = 4

10x + 7y = − 8

Solution: In this system, we cannot eliminate a variable by multiplying only one equation and then adding. To eliminate the variable x, we can multiply the first equation by 5 and the second equation by −3. We will be able to eliminate the x variable.


Continued, 10x + 7y = − 8 6x + 2y = 4

30x + 10y = 20 −30x − 21y = 24

− 11y = 44 y = − 4

We can now find x by substituting −4 for y in either of the original equations.

Substituting: 6x + 2y = 4

6x + 2(−4) = 46x − 8 = 4

6x = 12x = 2

The solution is (2, −4).


7.3

Matrices


Matrix

A matrix is a rectangular array of elements. An array is a systematic arrangement of numbers or symbols in rows and columns.

Matrices (the plural of matrix) may be used to display information and to solve systems of linear equations.The numbers in the rows and columns of a matrix are called the elements of the matrix.


Dimensions of a Matrix

The dimensions of a matrix may be indicated with the notation , where r is the number of rows and s is the number of columns of a matrix. A matrix that contains the same number of rows and columns is called a square matrix.

Example:

2 5 7 4 0 23 3 square matrix 1 2 6 4 1 3

4 3 8 6 8 9x

−⎛ ⎞⎛ ⎞⎜ ⎟⎜ ⎟⎜ ⎟⎜ ⎟⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠

r s×


Addition and Subtraction of Matrices

Two matrices can only be added or subtracted if they have the same dimensions.The corresponding elements of the two matrices are either added or subtracted.


Adding Matrices

Example: Find A + B.

Solution: A + B

5 7 1 0,

2 8 6 4A B⎛ ⎞ ⎛ ⎞= =⎜ ⎟ ⎜ ⎟−⎝ ⎠ ⎝ ⎠

5 7 1 02 8 6 45 1 7 0 6 72 6 8 4 4 12

⎛ ⎞ ⎛ ⎞= +⎜ ⎟ ⎜ ⎟−⎝ ⎠ ⎝ ⎠

+ +⎛ ⎞ ⎛ ⎞= =⎜ ⎟ ⎜ ⎟− + +⎝ ⎠ ⎝ ⎠


Multiplication of Matrices

A matrix may be multiplied by a real number, a scalar, by multiplying each entry in the matrix by the real number.Multiplication of matrices is possible only when the number of columns in the first matrix is the same as the number of rows of the second matrix. In general

.a b e f ae bg af bh

A Bc d g h ce dg cf dh

+ +⎛ ⎞⎛ ⎞ ⎛ ⎞× = =⎜ ⎟⎜ ⎟ ⎜ ⎟+ +⎝ ⎠⎝ ⎠ ⎝ ⎠


Example: Multiplying Matrices

Solution:

Find , given3 5 2 6 4

and 4 1 0 3 7

A B

A B

×

−⎛ ⎞ ⎛ ⎞= =⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

3 5 2 6 44 1 0 3 73( 2) 5(0) 3(6) 5(3) 3(4) 5(7)4( 2) 1(0) 4(6) 1(3) 4(4) 1(7)1 33 477 27 23

A B−⎛ ⎞⎛ ⎞

× = ⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠

− + + +⎛ ⎞= ⎜ ⎟− + + +⎝ ⎠

−⎛ ⎞= ⎜ ⎟−⎝ ⎠


Example: Identity Matrix in Multiplication

Use the multiplicative identity matrix for a 2 2 matrix and matrix A to show that

Solution: The identity matrix is

×

.A I A× =1 3

A 4 6⎡ ⎤

= ⎢ ⎥⎣ ⎦

1 0.

0 1I ⎛ ⎞= ⎜ ⎟⎝ ⎠


Example: Identity Matrix in Multiplication continued

1 3 1 0I

4 6 0 11(1) 3(0) 1(0) 3(1)4(1) 6(0) 4(0) 6(1)1 3

A4 6

A ⎡ ⎤ ⎡ ⎤× = ⎢ ⎥ ⎢ ⎥

⎣ ⎦ ⎣ ⎦+ +⎡ ⎤

= ⎢ ⎥+ +⎣ ⎦⎡ ⎤

= =⎢ ⎥⎣ ⎦


Multiplicative Identity Matrix

Square matrices have a multiplicative identity matrix. The following are the multiplicative identity for a 2 by 2 and a 3 by 3 matrix.

For any square matrix , .A A I I A A× = × =

1 0 01 0

0 1 00 1

0 0 1I I

⎛ ⎞⎛ ⎞ ⎜ ⎟= =⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎜ ⎟

⎝ ⎠


7.4

Solving Systems of Equations by Using Matrices


Augmented Matrix

The first step in solving a system of equations using matrices is to represent the system of equations with an augmented matrix.

An augmented matrix consists of two smaller matrices, one for the coefficients of the variables and one for the constants.

Systems of equations Augmented Matrixa1x + b1y = c1a2x + b2y = c2 1 1 1

2 2 2

a b ca b c⎛ ⎞⎜ ⎟⎝ ⎠


Row Transformations

To solve a system of equations by using matrices, we use row transformations to obtain new matrices that have the same solution as the original system.We use row transformations to obtain an augmented matrix whose numbers to the left of the vertical bar are the same as the multiplicative identity matrix.


Procedures for Row Transformations

Any two rows of a matrix may be interchanged.All the numbers in any row may be multiplied by any nonzero real number.All the numbers in any row may be multiplied by any nonzero real number, and these products may be added to the corresponding numbers in any other row of numbers.


Example: Using Row Transformations

Solve the following system of equations by using matrices.x + 2y = 162x + y = 11

Solution: First we write the augmented matrix.

Our goal is to obtain a matrix in the form:

It is generally easier to work in columns. Since the element in the top left position is already 1, we work to change the 2 in the first column second row to a zero.

1 2 162 1 11⎛ ⎞⎜ ⎟⎝ ⎠

1

2

1 00 1

cc

⎛ ⎞⎜ ⎟⎝ ⎠


Example: Using Row Transformations continued

If we multiply the top row of numbers by –2 and add these products to the second row of the second numbers, the element in the first column, second row will become 0.

The top row of numbers multiplied by –2 gives

1(−2), 2(−2) 16(−2)

Add these products to their respective numbers in row 2.

The next step is to obtain a 1 in the second column, second row. Multiplying all numbers in the second row by

1 2 160 3 21⎛ ⎞⎜ ⎟− −⎝ ⎠

1 .3

−

1 2 160 1 7⎛ ⎞⎜ ⎟⎝ ⎠


Example: Using Row Transformations continued

The next step is to obtain a 0 in the second column first row. Multiply the numbers in the second row by −2 and add these products to the corresponding numbers in the first row gives us 0.

With this desired augmented matrix, we see that our solution to the system is (2, 7).

This can be checked by substituting these values into the original equations.

1 0 20 1 7⎛ ⎞⎜ ⎟⎝ ⎠


Inconsistent and Dependent Systems

An inconsistent system occurs when obtaining an augmented matrix, one row of numbers on the left side of the vertical line are all zeros but a zero does not appear in the same row on the right side of the vertical line.

This indicates that the system has no solution.If a matrix is obtained and a 0 appears across an entire row, the system of equations is dependent.


Triangularization Method

The triangularization method can be used to solve a system of two equations.The ones and the zeros form a triangle.

In the previous problem we obtained the matrix

The matrix represents thefollowing equations.

x + 2y = 16y = 7

Substituting 7 for y in the equation then solving for x, x = 2.

10 1

a bc

⎛ ⎞⎜ ⎟⎝ ⎠

1 2 160 1 7⎛ ⎞⎜ ⎟⎝ ⎠


7.5

Systems of Linear Inequalities


Procedure for Solving a System of Linear Inequalities

Select one of the inequalities. Replace the inequality symbol with an equals sign and draw the graph of the equation. Draw the graph with a dashed line if the inequality is < or > and with a solid line if the inequality is .

Select a test point on one side of the line and determine whether the point is a solution to the inequality.

If so, shade the area on the side of the line containing the point. If the point is not a solution, shade the area on the other side of the line.

or ≤ ≥


Procedure for Solving a System of Linear Inequalities continued

Repeat steps 1 and 2 for the other inequality.The intersection of the two shaded areas and any solid line common to both inequalities form the solution set to the system.


Example: Solving a System of Inequalities

Graph the following system of inequalities and indicate the solution set.

Solution: Graph both inequalities on the same axis. The first graph will have a dashed line and the second graph a solid line.

12 1

x yy x+ >≤ +


Example: Another System of Inequalities

Graph the following system of inequalities and indicate the solution set.

Solution: Graph both on the same axes. The solution set is the region of the graph that is shaded in both colors. The point of intersection is not part of the solution because it does not satisfy the inequality y<1.

31

xy≥ −<


Example: Another System of Inequalities continued

13x

y≥ −<


7.6

Linear Programming


Linear Programming

In a linear programming problem, there are restrictions called constraints.

Each is represented as a linear inequality.The list forms a system of linear inequalities.

When a system of inequalities is graphed, often a feasible region is obtained.

The points where two or more boundaries intersect are called the vertices of the feasible region.


Fundamental Principle of Linear Programming

If the objective function, K = Ax + By, is evaluated at each point in a feasible region, the maximum and minimum values of the equation occur at vertices of the region.


Solving a Linear Programming Problem

Determine all necessary constraintsDetermine the objective function.Graph the constraints and determine the feasible region.Determine the vertices of the feasible region.Determine the value of the objective function at each vertex. The solution is determined by the vertex that yields the maximum or minimum value of the objective function.


Example

Planer Carpentry makes rocking horses and rocking airplanes. Each rocking horse requires 5 hours of woodworking and 4 hours of finishing. Each airplane requires 10 hours of woodworking and 3 hours of finishing. Each month Planer Carpentry has 600 hours of labor for woodworking and 140 hours for finishing. The profit on each rocking horse is $40 and on each airplane is $75. How many of each product should be made in order to maximize profit?


Example continued

Constraintsx = number of rocking horsesy = number of rocking airplanesP = 40x + 75y (profit function)5x + 10y ≤ 10 (woodworking hours)4x + 3y ≤ 240 (finishing hours)x ≥ 0 and y ≥ 0


Example continued

P = 40(24) + 75(48) = 4560(24, 48)

P = 40(0) + 75(60) = 4500(0, 60)P = 40(60) + 75(0) = 2400(60, 0)P = 40(0) + 75(0) = 0(0, 0)P = 40x + 75yVertices

The maximum profit would be from making 24 rocking horses and 75 rocking airplanes.

copyright seventh edition and expanded seventh …€¦ · · 2009-01-05copyright © 2005 pearson...

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